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  • mbenkerumass 6:00 am on May 29, 2020 Permalink | Reply
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    Sinusoidal and Exponential Sequences, Periodicity of Sequences 

    Continuing our discussion on discrete-time sequences, we now come to define exponential and sinusoidal sequences. The general formula for a discrete-time exponential sequence is as follows:

    x[n] = Aα^n.

    This exponential behaves differently according to the value of α. If the sequence starts at n=0, the formula is as follows:

    x[n] = Aα^n * u[n].


    If α is a complex number, the exponential function exhibits newer characteristics. The envelope of the exponential is |α|. If |α| < 1, the system is decaying. If |α|> 1, the system is growing.


    When α is complex, the sequence may be analyzed as follows, using the definition of Euler’s formula to express a complex relationship as a magnitude and phase difference.

    Captu56 ma

    Where ω0 is the frequency and φ is the phase, for n number of samples, a complex exponential sequence of form Ae^jw0n may be considered as a sinusoidal sequence for a set of frequencies in an interval of 2π.

    A sinusoidal sequence is defined as follows:

    x[n] = Acos(ω0*n + φ), for all n, and A, φ are real constants.

    Periodicity for discrete-time signals means that the sequence will repeat itself for a certain delay, N.

    x[n] = x[n+N] : system is periodic.

    t = (-5:1:15)’;

    impulse = t==0;
    unitstep = t>=0;
    Alpha1 = -0.5;
    Alpha2 = 0.5;
    Alpha3 = 2.5;
    Alpha4 = -2.5;
    cAlpha1 = -0.5 – 0.5i;
    cAlpha2 = 0.5 + 0.5i;
    cAlpha3 = 2.5 -2.5i;
    cAlpha4 = -2.5 + 2.5i;
    A = 1;

    Exp1 = A.*unitstep.*Alpha1.^t;
    Exp2 = A.*unitstep.*Alpha2.^t;
    Exp3 = A.*unitstep.*Alpha3.^t;
    Exp4 = A.*unitstep.*Alpha4.^t;

    cExp1 = A.*unitstep.*cAlpha1.^t;
    cExp2 = A.*unitstep.*cAlpha2.^t;
    cExp3 = A.*unitstep.*cAlpha3.^t;
    cExp4 = A.*unitstep.*cAlpha4.^t;

    stem(t, impulse)

    stem(t, unitstep)
    title(‘Unit Step’)
    stem(t, cExp1)
    title(‘Exponential: alpha = -0.5 – 0.5i’)

    stem(t, cExp2)
    title(‘Exponential: alpha = 0.5 + 0.5i’)

    stem(t, cExp3)
    title(‘Exponential: alpha = 2.5 -2.5i’)

    stem(t, cExp4)
    title(‘Exponential: alpha = -2.5 + 2.5i’)
    stem(t, Exp1)
    title(‘Exponential: alpha = -0.5’)

    stem(t, Exp2)
    title(‘Exponential: alpha = 0.5’)

    stem(t, Exp3)
    title(‘Exponential: alpha = 2.5’)

    stem(t, Exp4)
    title(‘Exponential: alpha = -2.5’)



  • mbenkerumass 6:00 am on May 27, 2020 Permalink | Reply
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    Discrete-Time Impulse and Unit Step Functions 

    Discrete-Time Signals are understood as a set or sequence of numbers. These sequences possess magnitudes or values at a given index.

    One mark of Discrete-Time Signals is that the index value is an integer. Thus, the sequence will have a magnitude or value for a whole number index such as -5, -4, 0, 6, 10000, etc.

    A discrete-time signal represented as a sequence of numbers takes the following form:

    x[n] = {x[n]},          -∞ < n < ∞,

    where n is any real integer (the index).

    An analog representation describes values of a signal at time nT, where T is the sampling period. The sampling frequency is the inverse of the sampling period.

    x[n] = X_a(nT),      -∞ < n < ∞.


    Common Sequences

    Both a very simple and important sequence is the unit sample sequence, “discrete time impulse” or simply “impulse,” equal to 1 only at zero and equal to zero otherwise.


    The discrete time impulse is used to describe an entire system using a delayed impulse. An entire sequence may also be shifted or delayed using the following relation:

    y[n] = x[n – n0],

    where n0 is an integer (which is the increment of indices by which the system is delayed. The impulse function delayed to any index and multiplied by the value of the system at that index can describe any discrete-time system. The general formula for this relationship is,


    The unit step sequence is related to the unit impulse. The unit step sequence is a set of numbers that is equal to zero for all numbers less than zero and equal to one for numbers equal and greater than zero.


    The unit step sequence is therefore equal to a sequence of delta impulses with a zero and greater delay.

    u[n] = δ[n] + δ[n-1] + δ[n-2] + . . .


    The unit impulse can also be represented by unit step functions:

    δ[n] = u[n] – u[n-1].

    Below I’ve plotted both the impulse and unit step function in matlab.


    t = (-10:1:10)';
    impulse = t==0;
    unitstep = t>=0;
    stem(t, impulse)
    stem(t, unitstep)
    title('Unit Step')



  • mbenkerumass 6:00 am on May 26, 2020 Permalink | Reply
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    Image Resolution 

    Consider that we are interested in building an optical sensor. This sensor contains a number of pixels, which is dependent on the size of the sensor. The sensor has two dimensions, horizontal and vertical. Knowing the size of the pixels, we will be able to find the total number of pixels on this sensor.

    The horizontal field of view, HFOV is the total angle of view normal from the sensor. The effective focal length, EFL of the sensor is then:

    Effective Focal Length: EFL = V / (tan(HFOV/2)),

    where V is the vertical sensor size in (in meters, not in number of pixels) and HFOV is the horizontal field of view. Horizontal field of view as an angled is halved to account that HFOV extends to both sizes of the normal of the sensor.

    The system resolution using the Kell Factor: R = 1000 * KellFactor * (1 / (PixelSize)),

    where the Pixel size is typically given and the Kell factor, less than 1 will approximate a best real case result and accounts for aberrations and other potential issues.

    Angular resolution: AR = R * EFL / 1000,

    where R is the resolution using the Kell factor and EFL is the effective focal length. It is possible to compute the angular resolution using either pixels per millimeter or cycles per millimeter, however one would need to be consistent with units.

    Minimum field of view: Δl = 1.22 * f * λ / D,

    which was used previously for the calculation of the spatial resolution of a microscope. The minimum field of view is exactly a different wording for the minimum spatial resolution, or minimum size resolvable.

    Below is a MATLAB program that computed these parameters, while sweeping the diameter of the lens aperture. The wavelength admittedly may not be appropriate for a microscope, but let’s say that you are looking for something in the infrared spectrum. Maybe you are trying to view some tiny laser beams that will be used in the telecom industry at 1550 nanometer.

    Pixel size: 3 um. HFOV: 4 degrees. Sensor size: 8.9mm x 11.84mm.


  • mbenkerumass 6:00 am on May 25, 2020 Permalink | Reply

    Spatial Resolution of a Microscope 

    Angular resolution describes the smallest angle between two objects that are able to be resolved.

    θ = 1.22 * λ / D,

    where λ is the wavelength of the light and D is the diameter of the lens aperture.

    Spatial resolution on the other hand describes the smallest object that a lens can resolve. While angular resolution was employed for the telescope, the following formula for spatial resolution is applied to microscopes.

    Spatial resolution: Δl = θf = 1.22 * f * λ / D,

    where θ is the angular resolution, f is the focal length (assumed to be distance to object from lens as well), λ is the wavelength and D is the diameter of the lens aperture.



    The Numerical Aperture (NA) is a measure of the the ability to of the lens to gather light and resolve fine detail. In the case of fiber optics, the numerical aperture applies to the maximum acceptance angle of light entering a fiber. The angle by the lens at its focus is θ = 2α. α is shown in the first diagram.

    Numerical Aperture for a lens: NA = n * sin(α),

    where n is the index of refraction of the medium between the lens and the object. Further,

    sin(α) = D / (2d).

    The resolving power of a microscope is related.

    Resolving power: x = 1.22 * d * λ / D,

    where d is the distance from the lens aperture to the region of focus.


    Using the definition of NA,

    Resolving power: x = 1.22 * d * λ / D = 1.22 * λ / (2sin(α)) = 0.61 * λ / NA.


  • mbenkerumass 6:00 am on May 24, 2020 Permalink | Reply

    Telescope Resolution & Distance Between Stars using the Rayleigh Limit 

    Previously, the Rayleigh Criterion and the concept of maximum resolution was explained. As mentioned, Rayleigh found this formula performing an experiment with telescopes and stars, exploring the concept of resolution. This formula may be used to determine the distance between two stars.

    θ = 1.22 * λ / D.

    Consider a telescope of lens diameter of 2.4 meters for a star of visible white light at approximately 550 nanometer wavelength. The distance between the two stars in lightyears may be calculated as follows. The stars are approximately 2.6 million lightyears away from the lens.

    θ = 1.22 * (550*10^(-9)m)/(2.4m)

    θ =2.80*10^(-7) rad

    Distance between two objects (s) at a distance away (r), separated by angle (θ): s = rθ

    s = rθ = (2.0*10^(6) ly)*(2.80*10^(-7)) = 0.56 ly.

    This means that the maximum resolution for the lens size, star distance from the lens and wavelength would be that two stars would need to be separated at least 0.56 lightyears for the two stars to be distinguishable.


  • mbenkerumass 6:00 am on May 23, 2020 Permalink | Reply

    Diffraction, Resolution and the Rayleigh Criterion 

    The wave theory of light includes the understanding that light diffracts as it moves through space, bending around obstacles and interfering with itself constructively and destructively. Diffraction grating disperses light according to wavelength. The intensity pattern of monochromatic light going through a small, circular aperture will produce a pattern of a central maximum and other local minima and maxima.


    The wave nature of light and the diffraction pattern of light plays an interesting role in another subject: resolution. The light which comes through the hole, as demonstrated by the concept of diffraction, will not appear as a small circle with sharply defined edges. There will appear some amount of fuzziness to the perimeter of the light circle.

    Consider if there are two sources of light that are near to each other. In this case, the light circles will overlap each other. Move them even closer together and they may appear as one light source. This means that they cannot be resolved, that the resolution is not high enough for the two to be distinguished from another.


    Considering diffraction through a circular aperture the angular resolution is as follows:

    Angular resolution: θ = 1.22 * λ/D,

    where λ is the wavelength of light, D is the diameter of the lens aperture and the factor 1.22 corresponds to the resolution limit formulated and empirically tested using experiments performed using telescopes and astronomical measurements by John William Strutt, a.k.a. Rayleigh for the “Rayleigh Criterion.” This factor describes what would be the minimum angle for two objects to be distinguishable.

  • mbenkerumass 6:00 am on May 22, 2020 Permalink | Reply

    Optical Polarizers in Series 

    The following problems deal with polarizers, which is a device used to alter the polarization of an optical wave.

    1. Unpolarized light of intensity I is incident on an ideal linear polarizer (no absorption). What is the transmitted intensity?

      Unpolarized light contains all possible angles to the linear polarizer. On a two dimensional plane, the linear polarizer will emit only that amount of light intensity that is found in the axis of polarization. Therefore, the Intensity of light emitted from a linear polarizer from incident unpolarized light will be half the intensity of the incident light.

    2. Four ideal linear polarizers are placed in a row with the polarizing axes vertical, 20 degrees to vertical, 55 degrees to vertical, and 90 degrees to vertical. Natural light of intensity I is incident on the first polarizer.

      a) Calculate the intensity of light emerging from the last polarizer.

      b) Is it possible to reduce the intensity of transmitted light (while maintaining some light transmission) by removing one of the polarizers?

      c) Is it possible to reduce the intensity of transmitted light to zero by removing a polarizer(s)?

      a) Using Malus’s Law, the intensity of light from a polarizer is equal to the incident intensity multiplied by the cosine squared of the angle between the incident light and the polarizer. This formula is used in subsequent calculations (below). The intensity of light from the last polarizer is 19.8% of the incident light intensity.

      b) My removing polarizer three, the total intensity is reduced to 0.0516 times the incident intensity.

      c) In order to achieve an intensity of zero on the output of the polarizer, there will need to exist an angle difference of 90 degrees between two of the polarizers. This is not achievable by removing only one of the polarizers, however it would be possible by removing both the second and third polarizer, leaving a difference of 90 degrees between two polarizers.





  • mbenkerumass 6:00 am on May 21, 2020 Permalink | Reply

    Jones Vector: Polarization Modes 

    The Jones Vector is a method of describing the direction of polarization of light. It uses a two element matrix for the complex amplitude of the polarized wave. The polarization of a light wave can be described in a two dimensional plane as the cross section of the light wave. The two elements in the Jones Vector are a function of the angle that the wave makes in the two dimensional cross section plane of the wave as well as the amplitude of the wave.


    The amplitude may be separated from the ‘mode’ of the vector. The mode of the vector describes only the direction of polarization. Below is a first example with a linear polarization in the y direction.


    Using the Jones Vector the mode can be calculated for any angle. See calculations below:


    The phase differences of the Jones Vector are plotted for a visual representation of the mode. If both components of the differ in phase, the plot depict a circular or oval pattern that intersects both components of the mode on a two dimensional plot. The simplest of plots to understand is a polarization of 90 degree phase difference. In this case, both magnitudes of the components of the mode will be 1 and a full circle is drawn to connect these points of the mode. In the case of a zero phase difference, this is demonstrated at 45 degrees where both sin(45deg) and cos(45deg) equal 0.707. In this case, the phase difference is plotted as a straight line, indicating that polarization is of equal phase from each axis of the phase difference plot.




  • mbenkerumass 6:00 am on May 18, 2020 Permalink | Reply

    Optical Polarization, Malus’s Law, Brewster’s Angle 

    In the theory of wave optics, light may be considered as a transverse electromagnetic wave. Polarization describes the orientation of an electric field on a 3D axis. If the electric field exists completely on the x-axis plane for example, light is considered to be polarized in this state.

    Non-polarized light, such as natural light may change angular position randomly or rapidly. The process of polarizing light uses the property of anisotropy and the physical mechanisms of dichroism or selective absorption, reflection or scattering. A polarizer is a device that utilizes these properties. Light exiting a polarizer that is linearly polarized will be parallel to the transmission axis of the polarizer.



    Malus’s law states that the transmitted intensity after an ideal polarizer is

    I(θ)=I(0)〖cos〗^2 (θ),

    where the angle refers to the angle difference between the incident wave and the transmission axis of the polarizer.

    Brewster’s Angle, an extension of the Fresnel Equation is a theory which states that the difference between a transmitted ray or wave into a material comes at a 90 degree angle to the reflected wave or ray along the surface. This situation is true only at the condition of the Brewster’s Angle. In the scenario where the Brewster’s Angle condition is met, the angle between the incident ray or wave and the normal, the reflected ray or wave and the surface normal and the transmitted ray or wave and the surface normal are all equal.


    If the Brewster’s Angle is met, the reflected ray will be completely polarized. This is also termed the polarization angle. The polarization angle is a function of the two surfaces.







  • mbenkerumass 6:00 am on May 17, 2020 Permalink | Reply

    Fourth Generation Optics: Thin-Film Voltage-Controlled Polarization 

    Michael Benker
    ECE591 Fundamentals of Optics & Photonics
    April 20,2020


    Dr. Nelson Tabiryan of BEAM Engineering for Advanced Measurements Co. delivered a lecture to explain some of the latest advances in the field of optics. The fourth generation of optics, in short includes the use of applied voltages to liquid crystal technology to alter the polarization effects of micrometer thin film lenses. Both the theory behind this type of technology as well as the fabrication process were discussed.


    First Three Generations of Optics

    A summary of the four generation of optics is of value to understanding the advancements of the current age. Optics is understood by many as one of the oldest branches of science. Categorized by applications of phenomena observable by the human eye, geometrical optics or refractive optics uses shape and refractive index to direct and control light.

    The second generation of optics included the use of graded index optical components and metasurfaces. This solved the issue of needing to use exceedingly bulky components although it would be limited to narrowband applications. One application is the use of graded index optical fibers, which could allow for a selected frequency to reflect through the fiber, while other frequencies will pass through.

    Anisotropic materials gave rise to the third generation of optics, which produced technologies that made use of birefringence modulation. Applications included liquid crystal displays, electro-optic modulators and other technologies that could control material properties to alter behavior of light.


    Fourth Generation Optics

    To advance technology related to optics, there are several key features needed for output performance. A modernized optics should be broadband, allowing many frequencies of light to pass. It should be highly efficient, micrometer thin and it should also be switchable. This technology is currently present.

    Molecule alignment in liquid crystalline materials is essential to the theory of fourth generation optics. Polarization characteristics of the lens is determined by molecule alignment. As such, one can build a crystal or lens that has twice the refractive index for light which is polarized in one direction. This device is termed the half wave plate, which polarizes light waves parallel and perpendicular to the optical axis of the crystal. Essentially, for one direction of polarization, a full period sinusoid wave is transmitted through the half wave plate, but with a reversed sign exit angle, while the other direction of polarization is allowed only half a period is allowed through. As a result of the ability to differentiate a sign of the input angle to the polarization axis (full sinusoid polarized wave), the result is an ability to alter the output polarization and direction of the outgoing wave as a function of the circular direction of polarization of the incident wave.

    The arrangement of molecules on these micrometer-thin lenses are not only able to alter the direction according to polarization, but also able to allow the lens to act as a converging lens or diverging lens. The output wave, a result of the arrangement of molecules in the liquid crystal lens has practically an endless number of uses and can align itself to behave as any graded index lens one might imagine. An applied voltage controls the molecular alignment.

    How does the lens choose which molecular alignment to use when switching the lens? The answer is that, during the fabrication process, all molecular alignments are prepared that the user plans on employing or switching to at some point. These are termed diffraction wave plates.



    Problem 1.


    The second lens is equivalent to the first (left) lens, rotated 180 degrees. In the case of a polarization-controlled birefringence application, one would expect lens 2 to exhibit opposite output direction for the same input wave polarization as lens 1. For lens 1 (left), clockwise circularly polarized light will exit with an angle towards the right, while counterclockwise circularly polarized light exits and an angle to the left. This is reversed for lens 2.



    Problem 2.


    There are as many states as there are diffractive waveplates. If there are six waveplates, then there will be 6 states to choose from.


  • mbenkerumass 6:00 am on May 16, 2020 Permalink | Reply
    Tags: LED,   

    LED Simulation in Atlas 

    This post features an LED structure simulated in ATLAS. The goal will be to demonstrate why this structure may be considered an LED. Light Emitting Diodes and Laser Diodes both serve as electronic-to-photonic transducers. Of importance to the operation of LEDs is the radiative recombination rate.

    The following LED structure is built using the following layers (top-down):

    • GaAs: 0.5 microns, p-type: 1e15
    • AlGaAs: 0.5 microns, p-type: 1e15, x=0.35
    • GaAs: 0.1 microns, p-type: 1e15, LED
    • AlGaAs: 0.5 microns, n-type: 1e18, x=0.35
    • GaAs: 2.4 microns, n-type: 1e18

    This structure uses alternating GaAs and AlGaAs layers.



  • mbenkerumass 6:00 am on May 15, 2020 Permalink | Reply

    Pulsed Lasers and Continuous-Wave Lasers 

    Continuous-Wave (CW) Lasers emit a constant stream of light energy. Power emitted is typically not very high, not exceeding killoWatts. Pulse Lasers were designed to produce much higher peak power output through the use of cyclical short bursts of optical power with intervals of zero optical power output. There are several important parameters to explore in relation the pulsed laser in particular.

    The period of the laser pulse Δt is the duration from the start of one pulse to the start of the next pulse. The inverse of the period Δt is the repetition rate or repetition frequency. The pulse width τ is calculated as the 3dB (half power) drop-off width.

    The Duty cycle is an important concept in signals and systems for periodic pulsed systems and is described as the ratio of the pulse duration to the duration of the period. Interestingly, the continuous wave lase can be considered as a pulse laser with 100% duty cycle.


    Power calculations and Pulse Energy remain as several important relations.

    • Average Power: the product of Peak pulsed power, repetition frequency and the pulse width
    • Pulsed Energy: Average power divided by the repetition frequency

    Other formulations of these parameters are found above.rereate


  • mbenkerumass 6:00 am on May 14, 2020 Permalink | Reply
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    Monochromaticity, Narrow Spectral Width and High Temporal & Spatial Coherence 

    A laser is a device that emits light through a process of optical amplification based on stimulated emission of electromagnetic radiation. A laser has high monochromaticity, narrow spectral width and high temporal coherence. These three qualities are interrelated, as will be shown.

    Monochromaticity is a term for a system, particularly in relation to light that references a constant frequency and wavelength. With the understanding that color is a result of frequency and wavelength, a monochromatic system also means that a single color is selected. A good laser will have only one output wavelength and frequency, typically referred to in relation to the wavelength (i.e. 1500 nanometer wavelength, 870 nanometer wavelength).

    A monochromatic system, made of only one frequency ideally is a single sinusoid function. A constant frequency sinusoid plotted in the frequency domain will have a line width approaching zero.


    The time τ that the wave behaves as a perfect sinusoid is related to the spectral line width. If the sinusoid takes an infinite time domain presence, the spectral line width is zero. The frequency domain plot in this scenario is a perfect pulse.

    If two frequencies are present in the time domain, the system is not monochromatic, which violates one of the principles of a perfect laser.


    Temporal Coherence is essentially a different perspective of the same relation present between monochromaticity and narrow spectral width. Coherence is the ability to predict the value of a system. Temporal coherence means that, given information related to the time of the system, the position or value of the system should be predictable. Given a sinousoid with a long time domain presence, the value of the sinusoid will be predictable given a time value. This is one condition of a proper laser.

    Spatial coherence takes a value of distance as a given. If the system is highly spatially coherent, the value of the system at a certain distance should predictable. This point is also a condition of a proper laser. This is also one differentiating point between a laser and an LED, since an LED’s light propagation direction is unpredictable at a certain time and certainly not in a certain distance. Light emitted from the LED may travel at any angle at any time. An LED does not produce coherent light; the Laser does.

  • mbenkerumass 6:00 am on May 13, 2020 Permalink | Reply
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    AlGaAs/GaAs Strip Laser 

    This project features a heterostructure semiconductor strip laser, comprised of a GaAs layer sandwiched between p-doped and n-doped AlGaAs. The model parameters are outlined below. The structure is presented, followed by output optical power as a function of injection current. Thereafter, contour plots are made of the laser to depict the electron and hole densities, recombination rate, light intensity and the conduction and valence band energies.




  • mbenkerumass 6:00 am on May 6, 2020 Permalink | Reply
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    Deriving Newton’s Lens Equation for a diverging lens 

    For a Diverging lens, derive a formula for the output angle with respect to the refractive indexes and input angle. Assume paraxial approximation and thin lens.



    For a Diverging lens, construct a derivation of Newton’s lens equation x_o*x_i = f^2.Ca

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